A) \[\frac{1}{{{t}_{3}}}=\frac{1}{{{t}_{2}}}-\frac{1}{{{t}_{1}}}\]
B) \[{{t}_{3}}^{2}={{t}_{1}}^{2}-{{t}_{2}}^{2}\]
C) \[{{t}_{3}}=\frac{{{t}_{1}}+{{t}_{2}}}{2}\]
D) \[{{t}_{3}}=\sqrt{{{t}_{1}}{{t}_{2}}}\]
Correct Answer: D
Solution :
When stone is thrown up \[h=u{{t}_{1}}+\frac{1}{2}g{{t}_{1}}^{2}\] ...(i) When thrown down\[h=u{{t}_{2}}+\frac{1}{2}g{{t}_{2}}^{2}\] ...(ii) When released\[h=\frac{1}{2}g{{t}_{3}}^{2}\] ...(iii) \[h{{t}_{2}}=-u\,{{t}_{1}}{{t}_{2}}+\frac{1}{2}g{{t}_{1}}^{2}{{t}_{2}}\] \[\frac{h{{t}_{1}}=+u\,{{t}_{1}}{{t}_{2}}+\frac{1}{2}g{{t}_{2}}^{2}{{t}_{1}}}{h({{t}_{2}}+\,{{t}_{1}})=+\frac{1}{2}g{{t}_{2}}{{t}_{1}}({{t}_{1}}+{{t}_{2}})}\] \[h=\frac{1}{2}g{{t}_{1}}{{t}_{2}}\] ?(iv) Comparing (iii) and (iv)\[{{t}_{3}}=\sqrt{{{t}_{1}}{{t}_{2}}}\]You need to login to perform this action.
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